因式分解 换元法 课件知识讲解

14.分解因式：x4+2005x 2+2004x+2005
解：设2005=a，
则原式 =x4+ax 2+(a-1)x+a =x4-x+ax 2+ax+a =x(x 3-1)+a(x 2+x+1) =(x 2+x+1)(x 2-x+a) =(x 2+x+1)(x 2-x+2005)
15.x(天y(津x竞y赛+1）)分+解(x因y式+：3)-2(x12+)y-+(x+y-12)
(x2+4x-3)(2x+4x+1)
12.(第13届五羊杯)分解因式:
(x4-4x2+1)(4x+32x+1)+140x
解：设 x4+1=a
则原式 =(a-4x2)(a+3x2)+10x4 =a2-x2a-12x4+10x4 =a2-x2a-2x4 =(a-2x2)(a+x2) =(x4-2x2+1)(x4+x2+1) =(x+1)2(x-1)2(x2+x+1)(x2-x+1)
换元法
1.(第12届五羊杯)分解因式:
(x4+x2-4)(x 4+x2+3)+10 =____________________.
解：设 x4+x2=y ，
则原式 =(y-4)(y+3)+10 =y2-y-2 =(y-2)(y+1) =(x4+x2-2)(x4+x2+1) =(x2+2)(x2-1)[(x2+1)2-x2] =(x+1)(x-1)2(+x2)(x2+x+1)(2x-x+1)
解:令x+y=a,xy=b
则原式=a(a+2b)+(b+1)(b-1) =a2+2ab+b2-1 =(a+b)2-1 =(a+b+1)(a+b-1) =(x+y+xy+1)(x+y+xy-1) =(x+1)(y+1)(xy+x+y-1)
8.(陕西中考)分解因式: (a2+a+1)(2a-6a+1)+122a
2.(第12届五羊杯）分解因式:
(x-2)3-(y-2)3-(x-y)3 =__________________.
解：设x-2=a，y-2=b，则x-y=a-b，
原式 =a3-b3 -(a-3b) =(a-b2)(+aab+2)b-(a-b2)-(2aab+2)b =3ab(a-b) =3(x-2)(y-2)(x-y)
(a-1)2(a2-3a+1)
9.(黄冈竞赛)分解因式:(2a+5)(a 2-9)(2a-7)-91
(a-4)(2a+7)(2a 2 -a-8)
10.(天津竞赛)分解因式:(x+1)(x+2)(x+3)(x+6)+x
2
(x2+6x+6)2
11.(1999年希望杯)分解因式(:x 2 -1)(x+3)(x+5)+12
13. (第14届五羊杯)分解因式:
(2x-33y+)(3x-23-y1)25(x3-y)
解:设2x-3y=a,3x-2y=b,则a+b=5(x-y)
原式 =a 3 +b 3 -(a+b) 3 =-3ab(a+b) =-3(2x-3y)(3x-2y)(5x-5y) =-15(x-y)(2x-3y)(3x-2y)
3.(2001年全国竞赛)若 x2+xy+y=14,y2+xy+x=28
则x+y的值为__________.
解：将两式相加，得 (x+y) 2来自百度文库(x+y)=42
即(x+y+7)(x+y-6)=0 ∴x+y+7=0 或 x+y-6=0 ∴x+y=-7或 x+y=6
4.(1999年武汉)设为实数,且满足
解:设a+b=x,则由已知可得 (a+b)[(a+b2)-3ab]+3ab-1=0
即
x3 -3abx+3ab-1=0
分解因式得 (x-1)(x 2+x+1-3ab)=0
所以x=1 或 (x-1)(x 2+x+1-3ab)=0
若 (x-1)(x 2+x+1-3ab)=0
即 a2 +b 2 -ab+a+b+1=0
配方,得 12[(a-b2)+(a+12+) (b+12])=0
所以a=b=-1,此时x=a+b=-2.
因此a+b=1或-2
17.(1998年安徽)证明:1997×1998×1999×2000+1 是一个整数的平方,并求出这个整数.
证明:设1997=a,则原式=a(a+1)(a+2)(a+3)+1
(x-1)3+1999(x-1)=-1
则x+y的值为(
(y-1)3+1999(y-1)=1
)
A.1 B. -1 C.2 D.-2
解:将两式相加,得(x-1) 3+(y-1) 3+1999(a+b)=0
令x-1=a,y-1=b,则x+y-2=a+b
∴ a 3 +b 3 +1999(a+b)=0
(a+b)(a2-ab+b2+1999)=0
∵ a2-ab+2b+199＞9³0 ∴a+b=0
5.(1999重庆）分解因式：1999x2-(19992-1)x-1999
(x-1999)(1999x+1)
6.(五羊杯初二)分解因式:
(6x-1)(2x-1)(3x-1)(x-1)+x
2
(6x2-6x+12)
7.（缙云杯竞赛）分解因式: (x+y)(x+y+2xy)+(xy+1)(xy-1)
解：设x+y=a，xy=b
则原式
1 =b(b+1)+b+3-2(a+2
)
-(a-1)2
=(b+1)2-a2
=(b+1+a)(b+1-a)
=(xy+x+y+1)(xy-x-y+1)
=(x+1)(y+1)(x-1)(y-1)
16.(2004年全国联赛)实数a,b满足 a3+b 3+3ab=1
求a+b的值.
=(a2+3a)(2a+3a+2)+1
=(a2+3a2)+2(a2+3a)+1
=(a2+3a+12)
=(19927+3 1997+21)
∵
1997 2+3× 1997+1=1997 2×000+1
=39940021